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Thread: Trignometry - modelling and problem solving

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    Trignometry - modelling and problem solving

    From a point A due north of a tower, the angle of elevation to the top of the tower is 45 degrees. From point B, 100M on a bearing of 120 DEGREES FROM A the angle of elevation is 26 degrees. Find the height of the tower.

    Please can you help with this problem
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    Re: Trignometry - modelling and problem solving

    From a point A due north of a tower, the angle of elevation to the top of the tower is 45 degrees. From point B, 100M on a bearing of 120 DEGREES FROM A the angle of elevation is 26 degrees. Find the height of the tower.

    Please can you help with this problem
    Trignometry - modelling and problem solving-castle_top.png

    From the image above Given: $\displaystyle \angle EAD = 45^\circ$, $\displaystyle \angle BAD = 120^\circ$ and $\displaystyle AB = 100m$

    We have to find the height of the tower $\displaystyle ED$


    Draw from point $\displaystyle G$ to $\displaystyle CD$, $\displaystyle GH$ which is perpendicular to $\displaystyle CD$ and to $\displaystyle BE$, $\displaystyle GO$ perpendicular to $\displaystyle BE$


    In $\displaystyle \triangle ABC$:
    $\displaystyle \begin{align*}\sin {\angle BAC} =& \frac{BC}{AB}\\ =& \frac{BC}{100}....\text{[because }AB = 100, \angle BAD = 120^\circ \text{ and }\angle BAC = 180^\circ - 120^\circ = 60^\circ\text{]}\end{align*}$
    $\displaystyle \therefore GH = BC = \frac{100\sqrt{3}}{2}$


    In $\displaystyle \triangle GAH$:
    $\displaystyle \sin {\angle GAH} = \frac{GH}{AG}$
    $\displaystyle \therefore AG = \frac{100\sqrt{3}}{\sqrt{2}}....\text{[} \angle GAH = 45^\circ \text{ and } GH = \frac{100\sqrt{3}}{2}}\text{]}$


    In $\displaystyle \triangle ABG$:
    $\displaystyle \begin{align*}BG =& \sqrt{AB^2 + AG^2 - 2*AB*AG*\cos {\angle BAG}}...\text{[ by cosine law]}\\=& 136.60...[AB = 100, AG = \frac{100\sqrt{3}}{\sqrt{2}}, \angle BAG = 75^\circ]\end{align*}$


    In $\displaystyle \triangle BOG$:
    $\displaystyle \sin{\angle OBG} = \frac{OG}{BG}$
    $\displaystyle \therefore OG = 59.88^\circ...[\angle OBG = 26^\circ \text{ and } BG = 136.60$


    In $\displaystyle \triangle BEF$:
    $\displaystyle \begin{align*}\angle FEB =& 180^\circ - \angle FBE - \angle BFE \\ =& 180^\circ - 26^\circ - 90^\circ\\=& 64^\circ\end{align*}$
    Now $\displaystyle \angle AGH = \angle GEF$ because $\displaystyle GF$ is parallel to $\displaystyle AD$ and straight line $\displaystyle AE$ intersects them.


    In $\displaystyle \triangle AGH$:
    $\displaystyle \angle AGH = 180^\circ - \angle GAH - \angle GHA = 45^\circ...[\angle GAH = 45^\circ, \angle GHA = 90^\circ$


    Now:
    $\displaystyle \angle FEB = \angle BEA + \angle AED$

    Because $\displaystyle \angle AED = \angle GEF = \angle AGH = 45^\circ\text{ and } \angle FEB = 64^\circ$:

    $\displaystyle \therefore \angle BEA = 19^\circ$



    In $\displaystyle \triangle OEG$:
    $\displaystyle \sin{\angle GEO} = \frac{OG}{GE}$

    Because $\displaystyle \angle BEA = 19^\circ \text{ and } OG = 59.88$
    $\displaystyle \therefore GE = 183.92$



    In $\displaystyle \triangle GEF$:
    $\displaystyle \cos{\angle GEF} = \frac{EF}{GE}$

    But $\displaystyle \angle GEF = 45^\circ \text{ and } GE = 183.92$
    $\displaystyle \therefore EF = 130.05$


    Now:
    $\displaystyle \begin{align*}ED = EF + FD = EF + BC =& 130.05 + \frac{100\sqrt{3}}{2}\\ =& 216.65\end{align*}$

    So the height of the tower is $\displaystyle 216.65 \text{ meter}$
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